RENSSELAER POLYTECHNIC INSTITUTE
TROY, NY
FINAL EXAM INTRODUCTION TO ENGINEERING ANALYSIS
(ENGR-1100)
NAME: Solution
Section: ___________
RIN:
_______________________________
Tuesday, May 14, 2012
Problem
Points
1
20
2
20
3
20
4
20
5
20
6
20
Total
100
Score
N.B.: You will be graded on 5 problems, 20 points per problem. Problems 1, 2, and 3 are mandatory
and will be graded. Before turning in your exam, please make sure you have circled the two problems
you want to be graded out of problems 4, 5 and 6.
Problem 1 (20 points)
Given the system of linear equations:
b)
-3x – y + 2z = 6.5
x
– 3z = -10.5
5x – 2y – z = 9.5
write the system of equations in a matrix form AX=B, and identify matrices: X,
A and B.
Calculate the determinant of matrix A using the method of cofactors.
(3)
(4)
c)
Determine adjoint of A (adj(A)).
(5)
d)
Determine the inverse of matrix A using the adjoint matrix method.
(3)
e)
Using Cramer’s law, find the unknown x.
(5)
a)
Note: You should show all work to receive full credit
−3 −1 2
= 1
0 −3 ,
5 −2 −1
det( ) = (−1)(−1)
( )=
=
(
(
=
det(
=
6.5
= −10.5 ,
=
9.5
[1 × −1— (−3) × 5) + (−2)(−1)
−6 −14 −2
= −5 −7 −11
3
−7
1
−6
−5
3
)=
= −14 −7 −7
−2 −11 1
−3/14
−5
3
1 −6
)=
=
−14 −7 −7 = −1/2
28
−1/14
−2 −11 1
−0.2143 −0.1786 0.1071
= −0.5000 −0.2500 −0.2500
−0.0714 −0.3929 0.0357
6.5
−1 2
−10.5 0 −3
9.5
−2 −1
( )
=
= = 1.5
) = 42
( )
−5/28
−1/4
−11/28
[−3 × −3 − 2 × 1] = 14 + 14 = 28
3/28
−1/4
1/28
Problem 2 (20 points)
A 40-kg packing crate is pulled by a rope as shown. The coefficient of static friction between the crate
and the floor is 0.35. If α = 40°, determine
(a) the magnitude of the force P required to move the crate,
(8)
(b) whether the crate will slide or tip.
(9)
(c) the friction force between the packing crate and the floor.
(3)
Draw necessary FBDs to receive full credits
Force P for which sliding is impending
(We assume that crate does not tip)
W  (40 kg)(9.81 m/s 2 )  392.4 N
Fy  0: N  W  P sin 40  0
N  W  P sin 40
(1)
Fx  0: 0.35 N  P cos 40  0
Substitute for N from Eq. (1):
P
0.35(W  P sin 40)  P cos 40  0
0.35W
0.35sin 40  cos 40
P  0.3532W 
P  138.6 N ◄(a)
Force P for which crate rotates about C
(We assume that crate does not slide)
M C  0: ( P sin 40)(0.8 m)  ( P cos 40)(0.5 m)
W (0.4 m)  0
P
0.4 W
 0.4458W
0.8sin 40  0.5cos 40
◄
Crate will first slide (b)
P  0.3532(392.4 N)
Ff=0.35N=0.35(392.4-138.6sin40o)
P  138.6 N ◄
Ff =106.16 N◄(c)
Problem 3 (20 points)
A force and a torque are applied to a frame consisting of two members: ABC and BD that are
connected through a smooth pin at B as shown.
(1)
(2)
(3)
(4)
(5)
Identify two-force member(s), if any, in the frame
Draw complete free-body-diagrams for members ABC and BD, respectively.
Write equilibrium equations for member ABC.
Determine the force at B on ABC in Cartesian vector form
Determine the reactions at C and D in Cartesian vector form
Solution:
(1) Member BD is a two-force member.
(2) FBDs:
(3) Below is one of several alternative forms.
(1)
(8)
(3)
(3)
(5)
∑F
= 0 ⇒ Cx + B
∑F
= 0 ⇒ Cy − B
x
y
∑M
C
3
32 +1.52
1.5
32 +1.52
=0
−150 = 0
3
= 0 ⇒ 150 × 3− B
(1)
3 +1.52
2
(2)
× 3− 300 = 0
(3)
(4) From (3),
B=
450 − 300
= 55.9 lb
3× 0.8944
B = 50i − 25j
(5) From (1)
3B
Cx = −
= −50
3.354
B = 50i − 25j
lb
lb
lb
From (2)
Cy = 150 + 55.9 × 0.4472 = 175
C = -50i + 175j lb
lb
Note on member BD, D = -B. The force B determined above is determined from member ABC so we
have:
D = B = 50i - 25j lb.
Problem 4 (20 points)
Considering the system shown in the figure, find the minimum and maximum
weight of box A to maintain the system at equilibrium.
 All the pulleys are frictionless, and box A is located in a vertical
frictionless guide.
 WB = 20 N, WC = 100 N, WD = 50 N.
37°
37°
Note: You need to show all work to receive full credit.
(1): ΣFy = 0 : T3=WD = 50 N
(2): ΣFy = 0 : 2T2 Sin(37)+T3-WC = 0
Maximum Weight:
(3) and (4) N1=0
T2 = 41.66 N
(4-a): ΣFy = 0 : T1=WA = 20 N
(3-a): ΣFy = 0 : T1 + T2 – WB = 0 → WB = 61.66
Problem 5 (20 points)
The figure shows a box being supported and held in place by means of three cables. The mass of
the box is 500 kg. At equilibrium:
a) Draw a complete, separate free body diagram showing the particle and all the forces
acting on it.
(5)
b) Write the equilibrium equations for the free body diagram
(4)
c) Determine the magnitude of the force in each of the three cables
(7)
d) Determine the angle between cables AC and CD
(4)
Problem 6 (20 points)
The figure below shows the horizontal view of a homogeneous stainless steel (density is 7800 kg/m 3)
object (shaded area). The thickness of the object is uniform and equal to 10 mm. Determine the
weight of the object and the location of the weight with respect to the origin of x-y coordinate system
shown in the figure.
(a) Problem picture
(b) Centroids and area of quadrant of a circle.
Solution:
The area and centroids for each shape is 1 point per shape, so totally 9 points;
Total area: 1 point
Each moment and total moment: 0.5 each; totally 4 points
Shape
Square
Circle
Quadrant
Total
Area [m2]
0.0576
-0.0113
-0.0113
0.035
xc,i [m]
0.12
0.1
0.189
My,i [m3]
0.006912
-0.00113
-0.002136
0.00365
yc,i [m]
0.12
0.08
0.189
Mx,i [m3]
0.006912
-0.0009
-0.002136
0.003872
The weight and centroids are (the weight is 2 points, xc and yc are each 2 points):
W = V rg = 0.035× 0.01× 7800 × 9.8 = 26.75N
My
0.00365
= 0.104m=104mm
A
0.035
M
0.003872
yc = x =
= 0.111m=111mm
A
0.035
xc =
=